cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A240926 a(n) = 2 + L(2*n) = 2 + A005248(n), n >= 0, with the Lucas numbers (A000032).

Original entry on oeis.org

4, 5, 9, 20, 49, 125, 324, 845, 2209, 5780, 15129, 39605, 103684, 271445, 710649, 1860500, 4870849, 12752045, 33385284, 87403805, 228826129, 599074580, 1568397609, 4106118245, 10749957124, 28143753125, 73681302249, 192900153620, 505019158609
Offset: 0

Views

Author

Kival Ngaokrajang, Aug 03 2014

Keywords

Comments

This sequence also gives the curvature of touching circles inscribed in a special way in the smaller segment of a circle of radius 5/4 cut by a chord of length 2.
Consider a circle C of radius 5/4 (in some length units) with a chord of length 2. This has been chosen so that the larger sagitta also has length 2. The smaller sagitta has length 1/2. The input, besides the circle C, is the circle C_0 with radius R_0 = 1/4, touching the chord and circle C. The following sequence of circles C_n with radii R_n, n >= 1, is obtained from the conditions that C_n touches (i) the circle C, (ii) the chord and (iii) the circle C_(n-1). The curvature of the n-th circle, C_n = 1/R_n, n >= 0, is conjectured to be a(n). See an illustration given in the link. As found by Wolfdieter Lang (see part II of the proof given by W. Lang in the link), this circle problem is related to the nonnegative solutions of the Pell equation X^2 - 5*Y^2 = 4: a(n) = 2 + X(n) = 2 + A005248(n). For the larger segment below the chord (with sagitta length 2) the sequence would be A115032, see W. Lang's proof given in part I of the link.
If the circle radius and the sagitta length were both equal to 1, the curvature sequence would be A099938.
Essentially a duplicate of A092387. - R. J. Mathar, Jul 07 2023

Links

Crossrefs

Cf. A000032, A005248, A005592, A099938, A115032, A246638, A246639, A246640, A246641, A246642, A338303.

Programs

  • Magma
    [2+Lucas(2*n): n in [0..40]]; // Vincenzo Librandi, Oct 08 2015
  • Mathematica
    Table[2 + LucasL[2 n], {n, 0, 50}] (* Vincenzo Librandi, Oct 08 2015 *)
  • PARI
    vector(100, n, n--; 2 + fibonacci(2*n-1) + fibonacci(2*n+1)) \\ Altug Alkan, Oct 08 2015

Formula

Conjectures (proved in the next entry) from Colin Barker, Aug 25 2014 (and Aug 27 2014): (Start)
a(n) = (2 + ((1/2)*(3-sqrt(5)))^n + ((1/2)*(3+sqrt(5)))^n).
a(n) = 4*a(n-1) - 4*a(n-2) + a(n-3).
G.f.: -(5*x^2-11*x+4) / ((x-1)*(x^2-3*x+1)). (End)
From Wolfdieter Lang, Aug 26 2014: (Start)
a(n) = 2 + S(n, 3) - S(n-2, 3) = 2 + 2*S(n, 3) - 3*S(n-1, 3).
a(n) = 3*a(n-1) - a(n-2) - 2, n >= 1, with a(-1)= 5 and a(0) = 4 (from the S(n, 3) recurrence or from A005248).
The first of the Colin Barker conjectures above is true because of the Binet-de Moivre formula for L(2*n) (see the Jul 24 2003 Dennis P. Walsh comment on A005248). With phi = (1+sqrt(5))/2, use 1/phi = phi-1, phi^2 = phi+1, (phi-1)^2 = 2 - phi.
His third conjecture (the g.f.) follows from the g.f. of A005248 by adding 2/(1-x).
His second conjecture (recurrence) with input a(-3) = 20, a(-2) = 9 and a(-1) = 5 (from the above given recurrence) leads to his g.f. with the expanded denominator. Thus all three conjectures are true. (End)
a(n) = A005592(n) + 3, with n > 0. - Zino Magri, Feb 16 2015
a(n) = (phi^n + phi^(-n))^2, where phi = A001622 = (1 + sqrt(5))/2. - Diego Rattaggi, Jun 10 2020
Sum_{k>=0} 1/a(k) = A338303. - Amiram Eldar, Oct 22 2020

Extensions

Edited: name changed (after proof has been given in part II of the W. Lang link), comments rewritten, cross refs. and link to Chebyshev index added. - Wolfdieter Lang, Aug 26 2014

A261995 The first of four consecutive positive integers the sum of the squares of which is equal to the sum of the squares of twenty-one consecutive positive integers.

Original entry on oeis.org

42, 123, 315, 1827, 4659, 13650, 34794, 201114, 512610, 1501539, 3827187, 22120875, 56382603, 165155802, 420955938, 2433095298, 6201573882, 18165636843, 46301326155, 267618362067, 682116744579, 1998054897090, 5092724921274, 29435586732234, 75026640329970
Offset: 1

Views

Author

Colin Barker, Sep 08 2015

Keywords

Comments

For the first of the corresponding twenty-one consecutive positive integers, see A261996.

Examples

			42 is in the sequence because 42^2 + ... + 45^2 = 7574 = 8^2 + ... + 28^2.

Links

Crossrefs

Cf. A157092, A246642, A261996.

Programs

  • PARI
    Vec(-3*x*(6*x^8+8*x^6+27*x^5-596*x^4+504*x^3+64*x^2+27*x+14)/((x-1)*(x^8-110*x^4+1)) + O(x^40))

Formula

G.f.: -3*x*(6*x^8+8*x^6+27*x^5-596*x^4+504*x^3+64*x^2+27*x+14) / ((x-1)*(x^8-110*x^4+1)).

A261996 The first of twenty-one consecutive positive integers the sum of the squares of which is equal to the sum of the squares of four consecutive positive integers.

Original entry on oeis.org

8, 44, 128, 788, 2024, 5948, 15176, 87764, 223712, 655316, 1670312, 9654332, 24607376, 72079892, 183720224, 1061889836, 2706588728, 7928133884, 20207555408, 116798228708, 297700153784, 872022648428, 2222647375736, 12846743269124, 32744310328592
Offset: 1

Views

Author

Colin Barker, Sep 08 2015

Keywords

Comments

For the first of the corresponding four consecutive positive integers, see A261995.

Examples

			8 is in the sequence because 8^2 + ... + 28^2 = 7574 = 42^2 + ... + 45^2.

Links

Crossrefs

Cf. A157092, A246642, A261995.

Programs

  • PARI
    Vec(4*x*(x^8+3*x^7+3*x^6+9*x^5-89*x^4-165*x^3-21*x^2-9*x-2)/((x-1)*(x^8-110*x^4+1)) + O(x^40))

Formula

G.f.: 4*x*(x^8+3*x^7+3*x^6+9*x^5-89*x^4-165*x^3-21*x^2-9*x-2) / ((x-1)*(x^8-110*x^4+1)).
Showing 1-3 of 3 results.

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